Question
How many distinct 5 digit numbers can be formed using the digits 3, 2, 3, 2, 4, 5.
Number of such numbers $=\frac{5 !}{2 !}+\frac{5 !}{2 !}=5 !=120$
Case II: Numbers are formed from 2, 2, 3, 3 and any one of 4 or 5
Number of such numbers $=\frac{5 !}{2 ! 2 !}+\frac{5 !}{2 ! 2 !}=60$
Required number of numbers = 120 + 60 = 180
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$\lim _{x \rightarrow 3}\left[\frac{\sqrt{2 x+6}}{x}\right]$